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Physics · LAWS OF MOTION
हिन्दीA block of mass \( m = 4\text{ kg} \) is placed on top of another block of mass \( M = 6\text{ kg} \), which rests on a smooth horizontal floor. The coefficient of static friction between the upper and lower blocks is \( \mu_s = 0.3 \). Taking \( g = 10\text{ m/s}^2 \), calculate the maximum horizontal force \( F_{\max} \) that can be applied to the lower block \( M \) such that both blocks accelerate together without relative slipping.
View Answer & Solution
Correct answer · Option A
\( 30\text{ N} \)
Explanation
Maximum acceleration of upper block without slipping is \( a_{\max} = \mu_s g = 0.3 \times 10 = 3\text{ m/s}^2 \). Maximum force applied to system \( F_{\max} = (M + m)a_{\max} = (6 + 4) \times 3 = 30\text{ N} \).
